ideal-gas equation and isoprocesses

The ideal-gas equation of state links pressure $p$, volume $V$, absolute temperature $T$, and amount of substance $\nu$:

$$pV=\nu RT,\qquad R\approx 8.31\ \mathrm{J\,mol^{-1}K^{-1}}.$$

In terms of the gas mass $m$ and molar mass $M$ it is written as $pV=\frac{m}{M}RT$. Temperature must always be substituted in kelvins ($T=t+273$) and pressure in pascals, so volume is conveniently expressed in m³. An isoprocess keeps one parameter ($p$, $V$, or $T$) constant while the other two obey a simple law. At $T=const$ Boyle–Mariotte's law holds, $pV=const$; at $p=const$ Gay-Lussac's law holds, $V/T=const$; at $V=const$ Charles's law holds, $p/T=const$. Moving between two states is easiest with the combined gas law $\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$.

A gas occupies a vessel of volume $V=2.0$ L at temperature $T=300$ K and pressure $p=150$ kPa. Treating the gas as ideal, find the amount of substance $\nu$.

A gas occupies a volume $V_1=5.0$ L at pressure $p_1=100$ kPa and temperature $T_1=280$ K. It is compressed to $V_2=2.0$ L while being heated to $T_2=350$ K. Find the final pressure $p_2$.

In a bicycle-pump cylinder, air occupies $V_1=100$ cm³ at pressure $p_1=100$ kPa and temperature $T_1=293$ K. During rapid compression the volume becomes $V_2=25$ cm³ and the air temperature rises to $T_2=340$ K. (a) Find the final pressure. (b) Compare it with the pressure for isothermal compression to the same volume and explain the difference.

Open the PhET «Gas Properties» simulation in ideal-gas mode and run three experiments, recording data in a table.

1. Isotherm. Fix $T$ and $\nu$, vary $V$, and measure $p$. Check that the product $pV$ stays constant.
2. Isobar. Maintain constant $p$ (movable piston with a load), vary $T$, and check that $V/T=const$.
3. Isochore. Fix $V$ (rigid vessel), vary $T$, and check that $p/T=const$.
   For each experiment, plot a graph in suitable coordinates. Safety: do not heat sealed vessels and do not use household pressurised cylinders — perform all observations in the simulation.

Show how the equation of state $pV=\nu RT$ reduces to simple laws when one parameter is constant and how these laws look on graphs.

A three-panel sketch:

• (a) $p$–$V$: isotherms are hyperbolas; the curve farther from the origin corresponds to a higher $T$;
• (b) $V$–$T$: isobars are straight lines whose extensions pass through the origin; the greater $p$, the smaller the slope;
• (c) $p$–$T$: isochores are straight lines through the origin; the smaller $V$, the greater the slope.

A scuba diver at a depth $h = 20\ \text{m}$ exhales a bubble of volume $V_1 = 0.50\ \text{cm}^3$. At that depth the absolute pressure is $p_1 = p_0 + \rho g h = 1.0\times10^5 + 1000\times9.8\times20 \approx 2.96\times10^5\ \text{Pa}$; the water temperature of about $10\ ^\circ\text{C}$ (283 K) stays the same all the way up, and at the surface the pressure is $p_0 = 1.0\times10^5\ \text{Pa}$.

This uses the isothermal branch of $pV = \nu RT$ — Boyle's law: with $T$ and $\nu$ fixed the product $pV$ is constant. Hence $V_2 = V_1 \dfrac{p_1}{p_2} = 0.50 \times \dfrac{2.96\times10^5}{1.0\times10^5} \approx 1.5\ \text{cm}^3$: the bubble grows almost threefold, and the buoyant force on it grows with its volume.

Before a trip the air in a car tyre is at $15\ ^\circ\text{C}$ (288 K) with an absolute pressure of $3.2\times10^5\ \text{Pa}$ (2.2 bar above atmospheric). After 30 minutes at $120\ \text{km/h}$ the tyre has warmed to $45\ ^\circ\text{C}$ (318 K), while its inner volume of about $25\ \text{l}$ is taken as unchanged.

This is the isochoric branch of $pV = \nu RT$ — Gay-Lussac's law: with $V$ and $\nu$ fixed the ratio $p/T$ is constant. So $p_2 = p_1 \dfrac{T_2}{T_1} = 3.2\times10^5 \times \dfrac{318}{288} \approx 3.5\times10^5\ \text{Pa}$, i.e. the gauge pressure rises by about 0.33 bar — the reason tyre pressures are checked on a cold tyre.

A hot-air balloon envelope holds $V_1 = 2000\ \text{m}^3$ of air at $20\ ^\circ\text{C}$ (293 K). The burner raises the temperature of the air inside to $120\ ^\circ\text{C}$ (393 K), and since the envelope is open at the bottom, the pressure inside stays equal to the outside pressure $1.0\times10^5\ \text{Pa}$.

This uses the isobaric branch of $pV = \nu RT$ — Charles's law: with $p$ fixed the ratio $V/T$ is constant. The air expands to $V_2 = V_1 \dfrac{T_2}{T_1} = 2000 \times \dfrac{393}{293} \approx 2680\ \text{m}^3$, and about $680\ \text{m}^3$ of hot air leaves through the bottom opening. The density falls from $1.20$ to $\dfrac{1.20 \times 293}{393} \approx 0.89\ \text{kg/m}^3$, so every cubic metre of envelope gives about $0.31\ \text{kg}$ of buoyant mass, that is roughly $3.0\ \text{N}$ of lift force.